Can the limit of a sequence of functions in $L_p$ and the limit in almost everywhere convergence be different?

Question

I'm sure this is an easy question, but I am somewhat confused. I'm considering here a measure space $(X, \mathcal{A}, \mu)$, and the functions are real valued.

Given a sequence of functions $(f_n) \in L_p$ that converges almost everywhere to a function $f$ and converges in $L_p$ to a function $g$, is it true that $f=g$ almost everywhere?

Does this answer your question? Does convergence in $L^p$ imply convergence almost everywhere? Is NOT the same question, but the answer there also answers this question (convergence on $L^p$ implies convergence a.e. of a subsequence). — Tito Eliatron, Jan 11 '21 at 19:44
I guess it doesn't, because I am assuming the $\lim f_n = f $ a.e., for some function $f$. — user480840, Jan 11 '21 at 19:46
Both types of convergence imply convergence in probability, and the limit in probability is unique (up to a null set). — Gabriel Romon, Jan 11 '21 at 19:49
If by convergence in probability, you mean convergence in measure - then it is not true that a.e. convergence implies convergence in measure. https://www.johndcook.com/blog/modes_of_convergence/ — user480840, Jan 11 '21 at 19:58

score 3 · Accepted Answer · answered Jan 11 '21 at 20:06

3

If $f_n \to g$ in $L^p$ there is some subsequence such that $f_{n_k}(x) \to g(x)$ for ae. $x$. Hence $g(x)=f(x)$ for ae. $x$.

answered Jan 11 '21 at 20:06

copper.hat

172,524

Can the limit of a sequence of functions in $L_p$ and the limit in almost everywhere convergence be different?

1 Answers1

Linked